1. **State the problem:** Solve the differential equation $$y^{(5)}+3y^{(4)}+y^{(3)}-y^{\prime\prime}-4y=0$$. 2. **Characteristic equation:** For linear differential equations with constant coefficients, we use the characteristic polynomial by replacing $y^{(n)}$ with $r^n$: $$r^5 + 3r^4 + r^3 - r^2 - 4 = 0$$ 3. **Factor the polynomial:** Try to find rational roots using the Rational Root Theorem. Test $r=1$: $$1 + 3 + 1 - 1 - 4 = 0$$ So, $r=1$ is a root. 4. **Divide polynomial by $(r-1)$:** $$\frac{r^5 + 3r^4 + r^3 - r^2 - 4}{r-1} = r^4 + 4r^3 + 5r^2 + 4r + 4$$ 5. **Factor the quartic:** Try to factor $r^4 + 4r^3 + 5r^2 + 4r + 4$ as $(r^2 + ar + b)(r^2 + cr + d)$. Expanding: $$r^4 + (a+c)r^3 + (ac + b + d)r^2 + (ad + bc)r + bd = r^4 + 4r^3 + 5r^2 + 4r + 4$$ Equate coefficients: - $a + c = 4$ - $ac + b + d = 5$ - $ad + bc = 4$ - $bd = 4$ Try $b = 2$, $d = 2$ (since $bd=4$): Then: - $ac + 4 = 5 \Rightarrow ac = 1$ - $ad + bc = 4$ From $a + c = 4$ and $ac = 1$, solve for $a$ and $c$: The quadratic for $a$ is: $$a^2 - 4a + 1 = 0$$ Roots: $$a = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}$$ Then $c = 4 - a = 2 \mp \sqrt{3}$. Check $ad + bc = 4$: $$a \cdot 2 + b \cdot c = 2a + 2c = 2(a + c) = 2 \times 4 = 8 \neq 4$$ Try $b=4$, $d=1$: Then $bd=4$ correct. - $ac + 5 = 5 \Rightarrow ac = 0$ - $ad + bc = 4$ Since $ac=0$, either $a=0$ or $c=0$. If $a=0$, then $c=4$ (from $a+c=4$). Check $ad + bc = 4$: $$a d + b c = 0 \times 1 + 4 \times 4 = 16 \neq 4$$ If $c=0$, then $a=4$. Check $ad + bc = 4$: $$a d + b c = 4 \times 1 + 4 \times 0 = 4$$ This works. So factors are: $$(r^2 + 4r + 4)(r^2 + 1)$$ 6. **Factor further:** $$(r^2 + 4r + 4) = (r + 2)^2$$ 7. **Complete factorization:** $$ (r - 1)(r + 2)^2 (r^2 + 1) = 0$$ 8. **Find roots:** - $r = 1$ - $r = -2$ (double root) - $r = i$ and $r = -i$ 9. **General solution:** For real roots $r_1$, $r_2$ (double), and complex roots $\alpha \pm \beta i$, the solution is: $$y = C_1 e^{r_1 x} + C_2 e^{r_2 x} + C_3 x e^{r_2 x} + e^{\alpha x}(C_4 \cos \beta x + C_5 \sin \beta x)$$ Here: $$y = C_1 e^{x} + C_2 e^{-2x} + C_3 x e^{-2x} + C_4 \cos x + C_5 \sin x$$ **Final answer:** $$\boxed{y = C_1 e^{x} + C_2 e^{-2x} + C_3 x e^{-2x} + C_4 \cos x + C_5 \sin x}$$